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3 years ago

If A, B, and C are the vertices of a triangle, find AB + BC + CA.

Mathematics

1 answer:

iVinArrow [24]3 years ago

5 0

Answer:

Step-by-step explanation:

When drawing a triangle, note that the vectors can be drawn either in a clockwise or anticlockwise direction. For a triangle with vertices A, B and C, te vectors are AB, BC and CA

In vector notation CA = - AC (length CA is equal to AC but acting in opposite direction to AC)

and AB+BC = AC

To evaluate AB + BC + CA, we will substitute the vector expression above into the given expression and simplify as shown;

= (AB + BC) + CA

= AC + (-AC)

open the parenthesis

= AC-AC

= 0

<em>Hence the sum of the vector expression AB + BC + CA is a zero vector</em>

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If x2 = 30, what is the value of x?<br> A. ±60 B. ±15 C. ±square root of 30 D.±square root of 15

anygoal [31]

Answer:

C. ±square root of 30

Step-by-step explanation:

Apply the square root function to both sides of the equation:

$\sqrt{x^2}=\sqrt{30}\\\\|x|=\sqrt{30}\\\\x=\pm\sqrt{30}$

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The absolute value equation has two solutions. They match choice C.

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3 years ago

X-1<br> 1<br> X+4<br> +<br> 2x+1<br> =<br> x-2<br> 2x²-3x-2

kiruha [24]

Answer:

x = 2π3

Step-by-step explanation:

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cot(3x)cot(3x) , x=2π3

Hope this helps :)

8 0

3 years ago

Plzz look into the question in the attachements

iren2701 [21]

Given : Diameter of the right circular cone ==> 8 cm

It means : The Radius of the right circular cone is 4 cm (as Radius is half of the Diameter)

Given : Volume of the right circular cone ==> 48π cm³

We know that :

$\bigstar \ \ \boxed{\textsf{Volume of a right circular cone is given by : $\pi r^2\dfrac{h}{3}$}}$

where : r is the radius of the circular cross-section.

h is the height of the right circular cone.

Substituting the respective values in the formula, we get :

$\mathsf{\implies \pi \times (4)^2 \times \dfrac{h}{3} = 48\pi}$

$\mathsf{\implies 16 \times \dfrac{h}{3} = 48}$

$\mathsf{\implies \dfrac{h}{3} = 3}$

$\implies \boxed{\mathsf{h= 9 \ cm}}$

<u>Answer</u> : Height of the given right circular cone is 9 cm

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2 years ago

From a circular cylinder of diameter 10 cm and height 12 cm are conical cavity of the same base radius and of the same height is

Nataliya [291]

<h3>Volume of the remaining solid = 628 cm^2</h3>

<h3>Whole surface area = 659.4 cm^2</h3>

Step-by-step explanation:

Now, Given that:-

Diameter (d) = 10 cm

So, Radius (r) = 10/2 = 5cm

Height of the cylinder = 12cm.

$volume \: of \: the \: cylinder \: = \pi {r}^{2} h$

$= > \pi \times {5}^{2} \times 12 {cm}^{3} = 300\pi {cm}^{3}$

Radius of the cone = 5 cm.

Height of the cone = 12 cm.

$slant \: height \: of \: the \: cone \: = \sqrt{ {h}^{2} + \: {r}^{2} }$

$= > \sqrt{ {5}^{2}+{12}^{2} } cm \: = 13cm$

Volume of the cone = 1/3 *πr^2h

$= > \frac{1}{3} \pi \times {5}^{2} \times 12 {cm}^{3} = 100\pi {cm}^{3}$

therefore, the volume of the remaining solid

$= 300\pi {cm}^{3} - 100\pi {cm}^{3} \\ = 200 \times 3.14 {cm}^{3} = 628 {cm}^{3}$

Curved surface of the cylinder =

$2\pi \: rh \: = 2\pi \times 5 \times 12 {cm}^{2} \\ = 120\pi {cm}^{2} .$

$curved \: surface \: of \: the \: cone \: = \pi \: rl \\ = \pi \times 5 \times 13 {cm}^{2} \\ = 65\pi {cm }^{2} \\ area \: of \: (upper)circular \: base \: \\ of \: cylinder \: = \\ = \pi \: {r}^{2} = \pi \times {5}^{2}$